1. Value of m to obtain the collision theory relation for temperature dependency of rate constant is ________
a) 1
b) 0
c) -1
d) 0.5
Explanation: The general equation for temperature dependency is k=koTm \(e^{-\frac{E}{RT}} \)
If m = 0 → Arrhenius equation
m = 1 → transition state theory
m = 0.5 → Collision theory.
2. The reaction rate of a bimolecular reaction at 300K is 10 times the reaction rate at 150K. Calculate the activation energy using collision theory.
a) 4928 J/mol
b) 5164 J/mol
c) 3281 J/mol
d) 1296 J/mol
Explanation: T1 = 150K, T2 = 300K and k2 = 10k1
When we have two values of k and T
k1=ko\(\sqrt{T1}e^{-\frac{E}{RT1}}\) and k2=ko\(\sqrt{T2}e^{-\frac{E}{RT2}}\)
Modifying it gives
ln(k1)=ln(ko)+0.5ln(T1) – \((\frac{E}{R})\frac{1}{T1}\) and ln(k2)=ln(ko) – \((\frac{E}{R})\frac{1}{T2} \)
On further simplification we get ln\((\frac{k1}{k2}) = 0.5ln(\frac{T1}{T2}) \frac{E}{R} (\frac{1}{T1}-\frac{1}{T2}) \)
ln\((\frac{k1}{10k1})\)=0.5ln\((\frac{150}{300}) – \frac{E}{8.314}(\frac{1}{300} – \frac{1}{150}) \)
E = 4928 J/mol.
3. For the reaction -rA = \(\frac{1670[A][Po]}{6+CA} \frac{kmol}{ml.hr}\) write the units for constants 6 and 1670.
a) kmol/ml and 1/hr
b) 1/hr and 1/hr
c) kmol and kmol
d) data insufficient
Explanation: To get the final units
-rA = \(\frac{1670[\frac{kmol}{ml}][\frac{kmol}{ml}]}{6+\frac{kmol}{ml}} \)
The units of constants have to be 1/hr for 1670 and kmol/ml for 6.
4. Which theory is the basis for Arrhenius equation?
a) Collision theory
b) Kinetic theory of gases
c) Ideal gas law
d) Charles and Boyles law
Explanation: Arrhenius equation is based on Kinetic theory of gases.
5. Molecularity refers to an elementary reaction.
a) true
b) false
Explanation: Elementary reaction occurs in single reaction step with no intermediates and molecularity is the number of particles involved in the reaction.
6. Molecularity is ___________ value and order is ___________ value.
a) an integer, a fractional
b) a fractional, an integer
c) both fractional
d) both integers
Explanation: Order is empirical, hence can be a fractional value but molecularity refers to the mechanism and is applicable only for elementary reactions.
7. What is the relative rate of reaction for the given reaction.
2A + 9B → 7C + 4D
a) –rA = -9rB=7rC=–4rD
b) –rA =-rB/9=rC/7=rD/4
c) –rA =-rB/9=7rC=4rD
d) –rA = 9rB=-7rC=–4rD
Explanation: Relative rate of reaction can be obtained from the ratio of stoichiometric co -efficient.
8. The overall order of reaction for –rA = kCA0.7CB1.3 is _______
a) 0
b) 1
c) 2
d) 3
Explanation: Order w.r.t A is 0.7 and B is 1.3.Hence, Overall order is 2.
9. The units of rate constant of nth order is ___________
a) time-1
b) concentration(n-1)
c) time-1 concentration(1-n)
d) time-1 concentration(n-1)
Explanation: General expression for the rate constant is time-1 concentration(1-n).
10. The unit for concentration equilibrium constant for the following reaction is _________
2A + 9B ←→7C + 4D
a) \((\frac{mol}{dm^3})\)2
b) \((\frac{mol}{dm^3})\)9
c) \((\frac{mol}{dm^3})\)-4
d) \((\frac{mol}{dm^3})\)-3
Explanation: Concentration eqm constant for a reversible reaction is
aA + bB → cC + dD is
K = \(\frac{cC^c cD^d}{cA^a cB^b}\) unit is \((\frac{mol}{dm^3})\)c+d-a-b when it is applied to the above equation we get -3.